[leetcode] 17. Letter Combinations of a Phone Number

Given a digit string, return all possible letter combinations that the number could represent.

A mapping of digit to letters (just like on the telephone buttons) is given below.

Input:Digit string "23"
Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].

Note:

Although the above answer is in lexicographical order, your answer could be in any order you want.

这道题是求出一串数字按键可能代表的字符串，缅怀一下Nokia，题目难度为Medium。

题目本身比较简单，典型的递归回溯题目，具体就不详细说明了，对回溯法还不太熟悉的同学请查看八皇后题目（传送门），具体代码：

class Solution {
void getLetters(vector<string>& rst, string curRst, const string& digits, int pos) {
if(pos == digits.size()) {
rst.push_back(curRst);
return;
}

vector<char> letter(4, 'a');
int cnt = 0;
switch(digits[pos]) {
case '2': {
cnt = 3;
letter[0] = 'a';
letter[1] = 'b';
letter[2] = 'c';
break;
}
case '3': {
cnt = 3;
letter[0] = 'd';
letter[1] = 'e';
letter[2] = 'f';
break;
}
case '4': {
cnt = 3;
letter[0] = 'g';
letter[1] = 'h';
letter[2] = 'i';
break;
}
case '5': {
cnt = 3;
letter[0] = 'j';
letter[1] = 'k';
letter[2] = 'l';
break;
}
case '6': {
cnt = 3;
letter[0] = 'm';
letter[1] = 'n';
letter[2] = 'o';
break;
}
case '7': {
cnt = 4;
letter[0] = 'p';
letter[1] = 'q';
letter[2] = 'r';
letter[3] = 's';
break;
}
case '8': {
cnt = 3;
letter[0] = 't';
letter[1] = 'u';
letter[2] = 'v';
break;
}
case '9': {
cnt = 4;
letter[0] = 'w';
letter[1] = 'x';
letter[2] = 'y';
letter[3] = 'z';
break;
}
default: break;
}
for(int i=0; i<cnt; i++) {
getLetters(rst, curRst+letter[i], digits, pos+1);
}
}
public:
vector<string> letterCombinations(string digits) {
vector<string> rst;
if(digits.empty()) return rst;
getLetters(rst, "", digits, 0);
return rst;
}
};